TS EAMCET · Physics · Electrostatics

A positive charge \(Q\) is placed on a conducting spherical shell with inner radius \(R_1\) and outer radius \(R_2\). A particle with charge \(q\) is placed at the center of the spherical cavity. The magnitude of the electric field at a point in the cavity, a distance \(r\) from center is

A zero
B \(\frac{Q}{4 \pi \varepsilon_0 r^2}\)
C \(\frac{q}{4 \pi \varepsilon_0 r^2}\)
D \(\frac{(Q+Q)}{4 \pi \varepsilon_0 r^2}\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{Q}{4 \pi \varepsilon_0 r^2}\)

Step-by-step Solution

Detailed explanation

Total charge on shell \(=+0\) Here, total charge on shell is \(+Q\). So, the \((-Q)\) is on the inner surface of shell. Hence, electric field inside the conductor \(=0\) According the Gauss's law, \[ \oint E d s=\frac{Q}{\varepsilon_0} \]…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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